Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

public class utkarsh {

    InputStream is;
    PrintWriter out;
    
    long maxl = 4L * (long)1e18, mod = (long)1e9 + 7L;
    int p[], maxi = 2 * (int)1e9, sz, np = 15485900, modi = (int)1e9 + 7;
    boolean b[];    // = isPrime();
    
    void solve(){
        //Enter code here utkarsh
        //sz=sieve(); //sz = 1000001;
        //out.println(sz);
        int i, j, k, n, m, z;
        n = ni();
        m = ni();
        int a[][] = new int[n][m];
        for(i = 0; i < n; i++){
            for(j = 0; j < m; j++){
                a[i][j] = ni();
            }
        }
        z = 0;
        int c[] = new int[m];
        int r[] = new int[n];
        if(n < m){
        for(i = 0; i < n; i++){
            k = maxi;
            for(j = 0; j < m; j++){
                if(k > a[i][j]){
                    k = a[i][j];
                }
            }
            for(j = 0; j < m; j++){
                a[i][j] -= k;
            }
            r[i] = k;
            z += k;
        }
        for(j = 0; j < m; j++){
            k = maxi;
            for(i = 0; i < n; i++){
                if(k > a[i][j]){
                    k = a[i][j];
                }
            }
            for(i = 0; i < n; i++){
                a[i][j] -= k;
            }
            c[j] = k;
            z += k;
        }
        }else{
        for(j = 0; j < m; j++){
            k = maxi;
            for(i = 0; i < n; i++){
                if(k > a[i][j]){
                    k = a[i][j];
                }
            }
            for(i = 0; i < n; i++){
                a[i][j] -= k;
            }
            c[j] = k;
            z += k;
        }
        for(i = 0; i < n; i++){
            k = maxi;
            for(j = 0; j < m; j++){
                if(k > a[i][j]){
                    k = a[i][j];
                }
            }
            for(j = 0; j < m; j++){
                a[i][j] -= k;
            }
            r[i] = k;
            z += k;
        }
        }
        for(i = 0; i < n; i++){
            for(j = 0; j < m; j++){
                if(a[i][j] > 0){
                    out.println("-1");
                    return;
                }
            }
        }
        out.println(z);
        for(i = 0; i < n; i++){
            for(j = 0; j < r[i]; j++){
                out.println("row " + (i+1));
            }
        }
        for(i = 0; i < m; i++){
            for(j = 0; j < c[i]; j++){
                out.println("col " + (i+1));
            }
        }
    }
    
    int sieve(){ int i, j, n = np; b = new boolean[n]; for(i = 3; i < n; i += 2){ b[i] = true; } b[2] = true; for(i = 3; i * i <= n; i += 2){ if(b[i]){ for(j = 2 * i; j < n; j += i){ b[j] = false; } } } p = new int[n]; p[0] = 2; j = 1; for(i = 3; i < n; i += 2){ if(b[i]){ p[j++] = i; } } p=Arrays.copyOf(p,j); return j;
    }
    long modpow(long base, long exp, long modulus) { base %= modulus; long result = 1L; while (exp > 0) { if ((exp & 1)==1) result = (result * base) % modulus; base = (base * base) % modulus; exp >>= 1; } return result;
    }
    
    public static void main(String[] args) { new utkarsh().run();
    }
    void run(){ is = System.in; out = new PrintWriter(System.out); solve(); out.flush();
    }
    
    byte input[] = new byte[1024];
    int len = 0, ptr = 0;
    
    int readByte(){ if(ptr >= len){ ptr = 0; try{ len = is.read(input); }catch(IOException e){ throw new InputMismatchException(); } if(len <= 0){ return -1; } } return input[ptr++];
    }
    boolean isSpaceChar(int c){ return !( c >= 33 && c <= 126 ); 
    }
    int skip(){ int b = readByte(); while(b != -1 && isSpaceChar(b)){ b = readByte(); } return b;
    }
    
    char nc(){ return (char)skip();
    }
    String ns(){ int b = skip(); StringBuilder sb = new StringBuilder(); while(!isSpaceChar(b)){ sb.appendCodePoint(b); b=readByte(); } return sb.toString();
    }
    int ni(){ int n = 0,b = readByte(); boolean minus = false; while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')){ b = readByte(); } if(b == '-'){ minus = true; b = readByte(); } if(b == -1){ return -1; } while(b >= '0' && b <= '9'){ n = n * 10 + (b - '0'); b = readByte(); } return minus ? -n : n;
    }
    long nl(){ long n = 0L; int b = readByte(); boolean minus = false; while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')){ b = readByte(); } if(b == '-'){ minus = true; b = readByte(); } while(b >= '0' && b <= '9'){ n = n * 10 + (b - '0'); b = readByte(); } return minus ? -n : n;
    }
    double nd(){ return Double.parseDouble(ns());
    }
    float nf(){ return Float.parseFloat(ns());
    }
    int[] na(int n){ int a[] = new int[n]; for(int i = 0; i < n; i++){ a[i] = ni(); } return a;
    }
    char[] ns(int n){ char c[] = new char[n]; int i,b = skip(); for(i = 0; i < n; i++){ if(isSpaceChar(b)){ break; } c[i] = (char)b; b = readByte(); } return i == n ? c : Arrays.copyOf(c,i);
    }
}